Question

PQ and RS are two parallel chords of a circle whose centre is O and radius is 10 cm. If PQ = 16 cm and RS = 12 cm, find the distance between PQ and RS, if they lie.
(i) on the same side of the centre O.

(ii) on opposite of the centre O.

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Answer 1

(i)

Draw the perpendicular bisector OL and OM of PQ and RS respectively.

PQ $\parallel$ RS

OL and OM are in the same line.

O, L and M are collinear.

Join OP and OR.

In right triangle $\Delta$OLP,

${OP}^2$ = ${OL}^2$ + ${PL}^2$ [By Pythagoras Theorem]

${\left(10\right)}^2$ = ${OL}^2$ + ${\left(\frac{PQ}{2}\right)}^2$ [The perpendicular drawn from the centre of a circle to a chord bisects the chord]

100 = ${OL}^2$ + ${\left(\frac{16}{2}\right)}^2$

100 = ${OL}^2$ + ${\left(8\right)}^2$

${OL}^2$ = 36

OL = 6 cm

In right triangle $\Delta$OMR,

${OR}^2$ = ${OM}^2$ + ${RM}^2$ [By Pythagoras Theorem]

= ${OM}^2$ + ${\left(\frac{RS}{2}\right)}^2$ [The perpendicular drawn from the centre of a circle to a chord bisects the chord]

${\left(10\right)}^2$ = ${OM}^2$ + ${\left(\frac{RS}{2}\right)}^2$

${\left(10\right)}^2$ = ${OM}^2$ + ${\left(\frac{12}{2}\right)}^2$

${OM}^2$ = 100 - 36

OM = 8 cm

LM = OM - OL = 8 - 6 = 2 cm

Hence, the distance between PQ and RS, if they lie on the same side of the centre O, is 2 cm.

(ii)

Draw the perpendicular bisector OL and OM of PQ and RS respectively.

PQ $\parallel$ RS

OL and OM are collinear.

Join OP and OR.

In right triangle $\Delta$OLP,

${OP}^2$ = ${OL}^2$ + ${PL}^2$ [By Pythagoras Theorem]

${\left(10\right)}^2$ = ${OL}^2$ + ${\left(\frac{PQ}{2}\right)}^2$ [The perpendicular drawn from the centre of a circle to a chord bisects the chord]

100 = ${OL}^2$ + ${\left(\frac{16}{2}\right)}^2$

100 = ${OL}^2$ + ${\left(8\right)}^2$

${OL}^2$ = 36

OL = 6 cm

In right triangle $\Delta$OMR,

${OR}^2$ = ${OM}^2$ + ${RM}^2$ [By Pythagoras Theorem]

= ${OM}^2$ + ${\left(\frac{RS}{2}\right)}^2$ [The perpendicular drawn from the centre of a circle to a chord bisects the chord]

${\left(10\right)}^2$ = ${OM}^2$ + ${\left(\frac{RS}{2}\right)}^2$

${\left(10\right)}^2$ = ${OM}^2$ + ${\left(\frac{12}{2}\right)}^2$

${OM}^2$ = 100 - 36

OM = 8 cm

LM = OL + OM = 6 + 8 = 14 cm

Hence, the distance between PQ and RS, if they lie on the opposite side of the centre O, is 14 cm.